courseshub.world · Lives in Mathematics and Physics

Siméon Denis Poisson

Siméon-Denis Poisson

A course in two strands: the life of the boy from Pithiviers who became the central figure of French mathematical physics after Laplace, and the equation, bracket, kernel, spot and distribution that carry his name.

Born 21 June 1781, Pithiviers Died 25 April 1840, Sceaux Copley Medal Royal Society, 1832 Level advanced undergraduate → graduate
Fresnel diffraction behind an opaque disc (Fresnel number 5), computed from the formulas of Lesson 09. The dashed circle is the edge of the geometric shadow. Poisson predicted the bright centre spot and found it absurd.
Lesson 00

How to use this course

Siméon Denis Poisson wrote some three hundred works and left his name on more objects than almost any other scientist: an equation, a bracket, a ratio, a kernel, a summation formula, a spot of light and a probability distribution. He was the leading mathematical physicist in France between the death of Pierre-Simon Laplace and the rise of Augustin-Louis Cauchy’s generation, and for decades he decided who passed the examinations of the École Polytechnique.

His method was consistent. He took a physical problem, wrote down the equations as he believed nature imposed them, and solved them by analysis. Sometimes his physical assumptions were wrong, as in his molecular theory of elasticity and his resistance to the wave theory of light. His mathematics outlasted his physics.

Prerequisites for Part II
  • Multivariable calculus, including the divergence theorem.
  • Lagrangian or Hamiltonian mechanics at the level of a first course.
  • Fourier series and the Fourier transform.
  • A first course in probability (binomial distribution, expectation, variance).

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Poisson wrote in the language of his time; the statements here use modern notation and, where relevant, modern hypotheses.

Reference

Timeline

1781Born 21 June in Pithiviers, in the Orléanais, son of a former soldier who held a minor administrative post.
c. 1795Apprenticed to his uncle, a surgeon in Fontainebleau; gives it up.
1796Enters the École Centrale at Fontainebleau.
1798Enters the École Polytechnique in first place.
1800Leaves the École without the final examination, on the strength of his memoirs; appointed répétiteur.
1802Deputy professor at the École Polytechnique.
1806Professor, succeeding Joseph Fourier.
1808Astronomer at the Bureau des Longitudes. Memoir on the inequalities of the planetary mean motions.
1809Chair of rational mechanics at the new Faculty of Sciences in Paris. Memoir on the variation of arbitrary constants, with the Poisson bracket.
1811Traité de mécanique, first edition.
1812Memoir on the distribution of electricity on conductors. Elected to the physics section of the Institut.
1813Publishes \(\nabla^2V=-4\pi\rho\) for the potential inside attracting matter.
1815–16Examiner at the military school of Saint-Cyr, then at the École Polytechnique.
1817Marries Nancy de Bardi.
1818Fellow of the Royal Society. Sits on the Academy commission judging the diffraction prize; derives the bright spot from Augustin Fresnel’s theory.
1820Councillor of the University.
1823Summation formula; adiabatic law for gases.
1827Geometer at the Bureau des Longitudes, succeeding Laplace.
1829Memoir on the equilibrium and motion of elastic bodies.
1831Nouvelle théorie de l’action capillaire. Reports with Lacroix on the memoir of Évariste Galois.
1832Copley Medal of the Royal Society.
1833Second, much enlarged edition of the Traité de mécanique.
1835Théorie mathématique de la chaleur.
1837Recherches sur la probabilité des jugements. Made a peer of France.
1840Dies 25 April at Sceaux, near Paris, aged 58. Buried at Père-Lachaise.
Part I · Formation
Lesson 01

Pithiviers and Fontainebleau, 1781–1798

Siméon Denis Poisson was born on 21 June 1781 in Pithiviers, a small town south of Paris. His father had served as a soldier and later held a modest administrative post. When the Revolution came, the father rose in local government and became president of the district of Pithiviers. The family was not wealthy, and the father wanted his son to have a secure profession.

The first choice was medicine. The boy was sent to an uncle, a surgeon in Fontainebleau, as an apprentice. He lacked the manual coordination and the interest the work demanded, and he returned home. In 1796 he enrolled at the new École Centrale in Fontainebleau, one of the secondary schools created by the Revolution. There a mathematics teacher recognised his ability and encouraged him toward the competitive entrance examination of the École Polytechnique in Paris.

In 1798, aged seventeen, he took the examination and came first in the whole of France.

Lesson 02

The École Polytechnique, 1798–1806

The École Polytechnique, founded in 1794, was then the most concentrated school of mathematics in the world. Its teachers included Joseph-Louis Lagrange and, as examiner and patron, Pierre-Simon Laplace. Both noticed Poisson quickly and remained his supporters for life.

In his first year he wrote a memoir on finite differences that attracted the attention of Adrien-Marie Legendre, an unusual distinction for a student. In his final year he wrote on the theory of equations and Bézout’s elimination theorem. On the strength of this work he was allowed to leave in 1800 without taking the final examination, and he was immediately appointed répétiteur, a teaching assistant. He was deputy professor from 1802. When Joseph Fourier left for Grenoble as prefect of the Isère, Poisson succeeded him in 1806 as full professor.

He kept clear of politics. In 1804, when students wanted to publish an attack on Napoleon, he persuaded them not to, to protect the school. The regime took this as loyalty.

Lesson 03

Into the Institut, 1806–1816

In 1806 senior mathematicians backed Poisson for the mathematics section of the Institut, but no seat fell vacant. Positions came instead from elsewhere. In 1808 he became an astronomer at the Bureau des Longitudes, and in 1809 professor of rational mechanics at the newly founded Faculty of Sciences in Paris.

His research of these years was in celestial mechanics. A memoir of 1808 on the inequalities of the planetary mean motions extended a stability result of Lagrange and Laplace to the next order of approximation. Lagrange, then over seventy, was stimulated by it to return to the subject. In 1809 Poisson introduced, in a memoir on the variation of arbitrary constants, the bracket that bears his name (Lesson 05). His Traité de mécanique of 1811 became a standard text.

Electricity and election

The Institut set electricity as the subject of its prize for 1811–12, a choice made with Poisson in mind. Using Charles-Augustin de Coulomb’s inverse-square law, he computed how charge distributes itself over two conducting spheres in contact or at a distance, a hard problem he solved with series expansions. After the physicist Étienne-Louis Malus died in February 1812, Poisson was elected in his place to the physics section of the Institut. He was 30. In 1815 he became examiner at the military school of Saint-Cyr, and in 1816 examiner at the École Polytechnique, a post of great influence.

Part II · The mathematics and physics
Lesson 04

Planetary stability and variation of constants

For the general reader. Will the planets keep their orbits forever, or will their mutual pulls slowly push one into the Sun or out of the system? Laplace and Lagrange showed that, to a first approximation, the sizes of the orbits do not drift. Poisson’s first major paper, in 1808, pushed the proof one step further, and the method he used led him directly to his bracket.

Goals
  • Understand the method of variation of constants for perturbed motion.
  • State the Laplace–Lagrange–Poisson results on the semi-major axes.
  • Distinguish periodic, secular and mixed terms.

Variation of constants

Let \(H=H_0+H_1\), where the unperturbed motion under \(H_0\) is known explicitly: \(q=q(c,t)\), \(p=p(c,t)\), with \(2n\) constants \(c_1,\dots,c_{2n}\) (for a planet, the orbital elements). Invert to write each \(c_i\) as a function of \((q,p,t)\). Each \(c_i\) is constant under \(H_0\), so under the full Hamiltonian

\[ \frac{dc_i}{dt}=\{c_i,H_1\}=\sum_{j=1}^{2n}\{c_i,c_j\}\,\frac{\partial H_1}{\partial c_j}, \]

where \(H_1\) is now expressed through the \(c\)’s and \(t\). Lagrange showed in 1808–09 that the coefficients do not depend on \(t\), and wrote the equations with the inverse matrix (the Lagrange brackets). Poisson wrote them, in 1809, directly with the brackets \(\{c_i,c_j\}\) that now carry his name.

Secular terms

In the planetary problem, \(H_1\) is the perturbing function from the other planets. Expanded in powers of the masses, solutions contain periodic terms, which oscillate; secular terms, proportional to \(t\), which grow; and mixed terms such as \(t\cos(\alpha t+\beta)\), which oscillate with growing amplitude.

Theorem (Laplace 1773, Lagrange 1776; Poisson 1808)

In the planetary problem without exact commensurabilities of the mean motions, the semi-major axes contain no secular terms to first order in the masses (Laplace, Lagrange). To second order they contain no purely secular terms, only periodic and mixed terms (Poisson).

The first-order argument is short. The equation for the semi-major axis \(a\) is \(\dot a=\frac{2}{na}\frac{\partial R}{\partial\lambda}\), where \(\lambda\) is the mean longitude and \(R\) the perturbing function. \(R\) is a sum of cosines of integer combinations of the planets’ mean longitudes. Secular change can only come from terms whose argument does not rotate, and when there are no commensurabilities, those terms do not contain \(\lambda\) at all. So their \(\lambda\)-derivative vanishes. Spiru Haretu showed in 1878 that genuine secular terms appear at third order, so the series results do not settle stability. The question passed to Henri Poincaré and later to KAM theory.

Exercise 4.1For \(H_0=\tfrac12(p^2+q^2)\), write \(q=a\cos t+b\sin t\), \(p=-a\sin t+b\cos t\). Express \(a,b\) in terms of \((q,p,t)\) and show \(\{a,b\}=1\) for all \(t\).

Show solution
\(a=q\cos t-p\sin t\), \(b=q\sin t+p\cos t\). Then \(\{a,b\}=\frac{\partial a}{\partial q}\frac{\partial b}{\partial p}-\frac{\partial a}{\partial p}\frac{\partial b}{\partial q}=\cos t\cdot\cos t-(-\sin t)\sin t=1\), independent of \(t\), as Lagrange’s theorem says.

Exercise 4.2Add the perturbation \(H_1=\tfrac\varepsilon2q^2\). Write the equations for \(\dot a,\dot b\), average \(H_1\) over one period of \(t\), and solve the averaged equations. Compare with the exact frequency \(\sqrt{1+\varepsilon}\).

Show solution
\(\dot a=\{a,b\}\partial_bH_1=\partial_bH_1\), \(\dot b=\{b,a\}\partial_aH_1=-\partial_aH_1\). With \(H_1=\frac\varepsilon2(a\cos t+b\sin t)^2\), the average is \(\langle H_1\rangle=\frac\varepsilon4(a^2+b^2)\). So \(\dot a=\frac\varepsilon2b\), \(\dot b=-\frac\varepsilon2a\), giving \(a=A\cos(\varepsilon t/2)\), \(b=-A\sin(\varepsilon t/2)\) and \(q=A\cos\big((1+\tfrac\varepsilon2)t\big)\). This matches \(\sqrt{1+\varepsilon}=1+\frac\varepsilon2+O(\varepsilon^2)\). A naive expansion in \(\varepsilon\) instead produces the mixed term \(-\frac\varepsilon2At\sin t\), which grows without bound although the true motion is bounded.
Lesson 05

The Poisson bracket

For the general reader. Lesson 04 showed how Poisson wrote the drift of orbital elements with brackets. Detached from celestial mechanics, the bracket combines any two quantities of a mechanical system into a third. It is now the basic operation of Hamiltonian mechanics, and quantum mechanics replaced it with the commutator.

Goals
  • Define the Poisson bracket and write Hamilton’s equations with it.
  • State Poisson’s theorem on integrals of motion.
  • Compute angular momentum brackets.

Take canonical coordinates \(q_1,\dots,q_n,p_1,\dots,p_n\) and a Hamiltonian \(H(q,p,t)\). For functions \(f,g\) of \((q,p,t)\) define

\[ \{f,g\}=\sum_{i=1}^{n}\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right). \]

Hamilton’s equations \(\dot q_i=\partial H/\partial p_i,\ \dot p_i=-\partial H/\partial q_i\) are equivalent to

\[ \frac{df}{dt}=\{f,H\}+\frac{\partial f}{\partial t}\quad\text{for every }f,\qquad \{q_i,p_j\}=\delta_{ij},\quad\{q_i,q_j\}=\{p_i,p_j\}=0. \]

Poisson introduced the bracket in 1809, before William Rowan Hamilton’s formulation (1834–35), in the setting of Lagrange’s method of variation of constants (Lesson 04).

Theorem (Poisson, 1809)

If \(f\) and \(g\) are integrals of motion of a Hamiltonian system, that is \(df/dt=dg/dt=0\) along every trajectory, then \(\{f,g\}\) is also an integral of motion.

Proof. The bracket is bilinear, antisymmetric and satisfies the Jacobi identity \(\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0\). For time-independent \(f,g,H\), the Jacobi identity gives \(\{\{f,g\},H\}=\{f,\{g,H\}\}-\{g,\{f,H\}\}=0\). With explicit time dependence, add \(\partial_t\{f,g\}=\{\partial_tf,g\}+\{f,\partial_tg\}\) and use \(\partial_t f=-\{f,H\}\). Carl Gustav Jacob Jacobi, who proved the identity, pointed out in the 1840s how powerful Poisson’s theorem is.

In 1925 Paul Dirac observed that the quantum commutator \([\hat f,\hat g]\) corresponds to \(i\hbar\{f,g\}\), so that the canonical relations \(\{q,p\}=1\) become \([\hat q,\hat p]=i\hbar\).

Exercise 5.1With \(L_x=yp_z-zp_y\), \(L_y=zp_x-xp_z\), \(L_z=xp_y-yp_x\), show \(\{L_x,L_y\}=L_z\). If \(H\) is rotation invariant, what does Poisson’s theorem say about \(L_x\) and \(L_y\)?

Show solution
Only pairs \((q_i,p_i)\) contribute. Terms in \(z,p_z\): \(\frac{\partial L_x}{\partial z}\frac{\partial L_y}{\partial p_z}-\frac{\partial L_x}{\partial p_z}\frac{\partial L_y}{\partial z}=(-p_y)(-x)-(y)(p_x)=xp_y-yp_x\). Terms in \(x\) and \(y\) vanish because \(L_x\) has no \(x,p_x\) and \(L_y\) has no \(y,p_y\). So \(\{L_x,L_y\}=L_z\). If \(L_x\) and \(L_y\) are conserved, then so is \(L_z\): two components of angular momentum force the third.

Exercise 5.2For the harmonic oscillator \(H=\tfrac12(p^2+\omega^2q^2)\), compute \(\{q,H\}\) and \(\{p,H\}\), and show that \(A=p^2-\omega^2q^2\) is not conserved by computing \(\dot A\).

Show solution
\(\{q,H\}=p\), \(\{p,H\}=-\omega^2q\). Then \(\dot A=\{A,H\}=\frac{\partial A}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial A}{\partial p}\frac{\partial H}{\partial q}=(-2\omega^2q)(p)-(2p)(\omega^2q)=-4\omega^2qp\ne0\). The energy \(H\) itself satisfies \(\{H,H\}=0\).
Lesson 06

Potential theory and Poisson’s equation

For the general reader. Laplace had shown that the gravitational potential obeys a simple equation in empty space. Poisson saw that inside matter the equation changes, and that the change measures the density of the matter. The same equation governs electric charge, and it is the starting point for gravity in Newtonian physics and for electrostatics.

Goals
  • State and derive Poisson’s equation.
  • Compute the potential of a uniform ball.
  • Understand Poisson’s result on surface charge on conductors.

For a mass or charge density \(\rho\) with compact support, define the Newtonian potential

\[ V(\mathbf x)=\int_{\mathbb R^3}\frac{\rho(\mathbf y)}{|\mathbf x-\mathbf y|}\,d^3y. \]

Laplace showed in the 1780s that \(\nabla^2V=0\) outside the matter. Laplace had at times asserted the equation everywhere.

Theorem (Poisson, 1813)

If \(\rho\) is continuously differentiable with compact support, then \(V\) is twice continuously differentiable and

\[ \nabla^2V=-4\pi\rho \quad\text{everywhere in }\mathbb R^3. \]

With the sign convention \(\phi=-GV\) for gravity this reads \(\nabla^2\phi=4\pi G\rho\); in Gaussian electrostatics \(\nabla^2\phi=-4\pi\rho\), and in SI units \(\nabla^2\phi=-\rho/\varepsilon_0\). Poisson’s argument treated a small sphere around the point as uniform. The modern proof uses \(\nabla^2\frac1{|\mathbf x|}=-4\pi\delta\) in the sense of distributions: by Green’s identity, for a test function \(\varphi\), \(\int\frac{\nabla^2\varphi}{|\mathbf x|}\,d^3x=\lim_{\varepsilon\to0}\oint_{|\mathbf x|=\varepsilon}\big(\tfrac1{|\mathbf x|}\partial_n\varphi-\varphi\,\partial_n\tfrac1{|\mathbf x|}\big)dS=-4\pi\varphi(0)\), with \(n\) the normal pointing toward the origin. Rigorous proofs, and the hypothesis needed on \(\rho\) (Hölder continuity suffices, mere continuity does not), came later in the century.

Conductors

In his 1812 memoir Poisson treated a conductor in equilibrium: the potential is constant on it, all charge sits on the surface, and just outside the surface the field is normal with magnitude \(4\pi\sigma\), where \(\sigma\) is the surface charge density. He reduced the problem of two charged spheres to functional equations and solved them by series.

Exercise 6.1A ball of radius \(R\) has uniform density \(\rho_0\) and total mass \(M=\tfrac43\pi R^3\rho_0\). Show that \(V(r)=M/r\) for \(r\ge R\) and \(V(r)=\frac{M}{2R^3}(3R^2-r^2)\) for \(r\le R\), and verify Poisson’s equation inside.

Show solution
Outside, the shell theorem gives \(M/r\). Inside, Gauss’s law gives \(-V'(r)=M(r)/r^2=Mr/R^3\), so \(V=-\frac{Mr^2}{2R^3}+C\); continuity at \(R\) gives \(C=\frac{3M}{2R}\). For a radial function \(\nabla^2V=\frac1{r^2}(r^2V')'=\frac1{r^2}\big(-\frac{Mr^3}{R^3}\big)'=-\frac{3M}{R^3}=-4\pi\rho_0\).

Exercise 6.2Use the divergence theorem on a small pillbox across a charged surface to show that the jump in the normal component of \(\mathbf E=-\nabla\phi\) is \(4\pi\sigma\). Why is the field just outside a conductor \(4\pi\sigma\) and not \(2\pi\sigma\)?

Show solution
\(\nabla\cdot\mathbf E=4\pi\rho\). Integrating over a pillbox of face area \(A\) and vanishing height gives \((E_n^+-E_n^-)A=4\pi\sigma A\). Inside a conductor in equilibrium \(\mathbf E=0\), so \(E_n^+=4\pi\sigma\). An isolated thin sheet has symmetric fields \(\pm2\pi\sigma\); on a conductor the other charges cancel the field inside and double it outside.
Lesson 07

Fourier analysis: kernel and summation

For the general reader. Fourier claimed that any function can be written as a sum of sines and cosines. Poisson, sceptical of Fourier’s rigour, found a way to make such sums behave: damp each term slightly, then let the damping vanish. The damping produces the Poisson kernel, which solves the problem of the steady temperature in a disc. Another of his formulas trades a sum over integers for a sum of Fourier transforms.

Goals
  • Derive the Poisson kernel and solve the Dirichlet problem in the disc.
  • State and prove the Poisson summation formula.
  • Apply it to theta functions.

The Poisson kernel

Let \(f\) be continuous on the circle with Fourier coefficients \(c_n=\frac1{2\pi}\int_0^{2\pi}f(\varphi)e^{-in\varphi}d\varphi\). Instead of \(\sum c_ne^{in\theta}\), which may diverge, consider \(\sum r^{|n|}c_ne^{in\theta}\) for \(0\le r<1\). Summing the geometric series,

\[ \sum_{n\in\mathbb Z}r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos\theta+r^2}=:P_r(\theta). \]
Theorem (Poisson integral formula)

Let \(f\) be continuous on the unit circle. Then

\[ u(r,\theta)=\frac1{2\pi}\int_0^{2\pi}P_r(\theta-\varphi)\,f(\varphi)\,d\varphi \]

is harmonic in the open unit disc, and \(u(r,\theta)\to f(\theta)\) uniformly as \(r\to1^-\). It is the unique solution of the Dirichlet problem with boundary values \(f\).

\(P_r>0\), \(\frac1{2\pi}\int P_r=1\), and \(P_r\) concentrates at \(\theta=0\) as \(r\to1\); these three properties make it an approximate identity, exactly as the Gaussian in Weierstrass’s later proof of polynomial approximation. Harmonicity follows because each \(r^{|n|}e^{in\theta}\) is the real or imaginary part of \(z^n\) or \(\bar z^n\). Uniqueness follows from the maximum principle.

The summation formula (1823)

Theorem (Poisson summation formula)

Let \(f\in L^1(\mathbb R)\) with Fourier transform \(\hat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi ix\xi}dx\), and suppose \(|f(x)|+|\hat f(x)|\le C(1+|x|)^{-1-\delta}\) for some \(\delta>0\). Then

\[ \sum_{n\in\mathbb Z}f(n)=\sum_{k\in\mathbb Z}\hat f(k). \]

Proof. The periodization \(F(x)=\sum_nf(x+n)\) has period 1 and is continuous. Its Fourier coefficients are \(\int_0^1F(x)e^{-2\pi ikx}dx=\int_{\mathbb R}f(x)e^{-2\pi ikx}dx=\hat f(k)\), which are summable. So \(F(x)=\sum_k\hat f(k)e^{2\pi ikx}\), and \(x=0\) gives the formula. Poisson found it while studying the error term in the Euler–Maclaurin formula.

Exercise 7.1Apply the summation formula to \(f(x)=e^{-\pi tx^2}\), \(t>0\), to prove the theta transformation \(\theta(t)=t^{-1/2}\theta(1/t)\), where \(\theta(t)=\sum_ne^{-\pi n^2t}\).

Show solution
\(\hat f(\xi)=t^{-1/2}e^{-\pi\xi^2/t}\) (the Gaussian \(e^{-\pi x^2}\) is its own transform; rescale). Then \(\sum_ne^{-\pi n^2t}=t^{-1/2}\sum_ke^{-\pi k^2/t}\). Riemann used this identity in 1859 to prove the functional equation of the zeta function.

Exercise 7.2Verify the geometric series for \(P_r\): show \(\sum_{n\in\mathbb Z}r^{|n|}e^{in\theta}=\dfrac{1-r^2}{1-2r\cos\theta+r^2}\).

Show solution
With \(z=re^{i\theta}\), the sum is \(\frac1{1-z}+\frac{\bar z}{1-\bar z}=\mathrm{Re}\,\frac{1+z}{1-z}\). Multiply numerator and denominator by \(1-\bar z\): \(\frac{(1+z)(1-\bar z)}{|1-z|^2}=\frac{1-|z|^2+z-\bar z}{|1-z|^2}\), whose real part is \(\frac{1-r^2}{1-2r\cos\theta+r^2}\).

Exercise 7.3Use the summation formula with \(f(x)=e^{-2\pi a|x|}\), \(a>0\), to show \(\displaystyle\sum_{k\in\mathbb Z}\frac{1}{k^2+a^2}=\frac{\pi}{a}\coth(\pi a)\).

Show solution
\(\hat f(\xi)=\int e^{-2\pi a|x|-2\pi ix\xi}dx=\frac{a}{\pi(a^2+\xi^2)}\). The left side is \(\sum_ne^{-2\pi a|n|}=\frac{1+e^{-2\pi a}}{1-e^{-2\pi a}}=\coth(\pi a)\). So \(\coth(\pi a)=\frac a\pi\sum_k\frac1{k^2+a^2}\). Letting \(a\to0\) after subtracting the \(k=0\) term gives \(\sum_{k\ge1}1/k^2=\pi^2/6\).
Lesson 08

Waves, heat and elastic solids

For the general reader. How does a sound spread from a point in three dimensions? How does a gas cool when it expands? How much does a rubber band thin when you stretch it? Poisson answered all three with formulas still taught today. Not all his physical reasoning survived, but the formulas did.

Goals
  • Use Poisson’s formula for the wave equation in three dimensions and see Huygens’ principle.
  • Derive the adiabatic law \(pV^\gamma=\text{const}\).
  • Define Poisson’s ratio and understand his prediction \(\nu=\tfrac14\).

The wave equation

Theorem (Poisson’s formula)

Let \(g\in C^3(\mathbb R^3)\), \(h\in C^2(\mathbb R^3)\). The solution of \(u_{tt}=c^2\nabla^2u\) with \(u(\mathbf x,0)=g\), \(u_t(\mathbf x,0)=h\) is

\[ u(\mathbf x,t)=\frac{\partial}{\partial t}\Big(t\,M_{ct}[g](\mathbf x)\Big)+t\,M_{ct}[h](\mathbf x),\qquad M_s[\phi](\mathbf x)=\frac1{4\pi}\int_{|\boldsymbol\omega|=1}\phi(\mathbf x+s\boldsymbol\omega)\,dS(\boldsymbol\omega). \]

The key step is that for a radial solution \(u(r,t)\), the function \(ru\) satisfies the one-dimensional wave equation, which d’Alembert had solved. The value of \(u\) at \((\mathbf x,t)\) depends only on the data on the sphere of radius \(ct\) around \(\mathbf x\). A sharp pulse is therefore heard as a sharp pulse and then silence: this is Huygens’ principle in its strong form, which fails in two dimensions.

The adiabatic law (1823)

For an ideal gas compressed without heat exchange, Poisson derived \(pV^\gamma=\text{const}\), with \(\gamma=c_p/c_v\), the ratio of specific heats that Laplace had used to correct Newton’s speed of sound. Equivalent forms are \(TV^{\gamma-1}=\text{const}\) and \(T\propto p^{(\gamma-1)/\gamma}\).

Elasticity and Poisson’s ratio

When a bar is stretched by a small axial strain \(\varepsilon_{\text{ax}}\), it contracts sideways by \(\varepsilon_{\text{tr}}\). Poisson’s ratio is \(\nu=-\varepsilon_{\text{tr}}/\varepsilon_{\text{ax}}\). For an isotropic linear solid with Lamé constants \(\lambda,\mu\),

\[ \nu=\frac{\lambda}{2(\lambda+\mu)},\qquad E=\frac{\mu(3\lambda+2\mu)}{\lambda+\mu}. \]

In his 1829 memoir Poisson modelled a solid as molecules interacting by central forces. That model forces \(\lambda=\mu\), hence \(\nu=\tfrac14\) for every isotropic material. Measurements later showed that \(\nu\) varies (about 0.3 for steel, close to 0.5 for rubber), and the two-constant theory of Cauchy and George Green prevailed. The name of the ratio stayed with Poisson.

Exercise 8.1Derive \(pV^\gamma=\text{const}\) from the first law \(dU=-p\,dV\) for an adiabatic process, with \(U=c_vT\) per mole and \(pV=RT\), \(c_p-c_v=R\).

Show solution
\(c_v\,dT=-p\,dV=-\frac{RT}{V}dV\), so \(\frac{dT}{T}=-\frac{R}{c_v}\frac{dV}{V}=-(\gamma-1)\frac{dV}V\). Hence \(TV^{\gamma-1}=\text{const}\), and substituting \(T=pV/R\) gives \(pV^\gamma=\text{const}\).

Exercise 8.2Show that \(\lambda=\mu\) gives \(\nu=\tfrac14\). Using \(K=\lambda+\tfrac23\mu\) for the bulk modulus, show that \(K>0,\ \mu>0\) forces \(-1<\nu<\tfrac12\).

Show solution
\(\nu=\lambda/(4\lambda)=\tfrac14\). Writing \(\lambda=K-\tfrac23\mu\), \(\nu=\frac{3K-2\mu}{2(3K+\mu)}\). As \(\mu/K\) runs over \((0,\infty)\), this decreases from \(\tfrac12\) toward \(-1\). Materials with \(\nu<0\) (auxetics) exist; \(\nu\to\tfrac12\) means incompressible.

Exercise 8.3Use Poisson’s formula with \(g=0\) and \(h\) the indicator of the unit ball (ignore smoothness) to find \(u(\mathbf 0,t)\) for \(c=1\). For which times is it nonzero?

Show solution
\(M_t[h](\mathbf 0)=1\) if \(t<1\) and \(0\) if \(t>1\), since the sphere of radius \(t\) about the origin lies inside or outside the ball. So \(u(\mathbf0,t)=t\) for \(0\le t<1\) and \(0\) afterwards: the disturbance passes and leaves no trace, as Huygens’ principle predicts.
Lesson 09

The bright spot

For the general reader. If light is made of waves, the shadow of a round disc should have a bright point at its very centre, because waves from the rim all arrive there in step. Poisson, a believer in light as particles, worked this out from his opponent’s theory and thought it proved the theory absurd. The experiment was done, and the spot was there.

Goals
  • Set up Fresnel diffraction for a circular obstacle.
  • Compute the on-axis intensity exactly.
  • Understand the place of the episode in the history of optics.

The episode

The Academy of Sciences set diffraction as the subject of its physics prize, judged in 1818–19. Most of the commission, including Laplace, Jean-Baptiste Biot and Poisson, favoured the corpuscular theory of light. François Arago chaired it. Augustin Fresnel submitted a wave theory based on Huygens’ principle with interference. Poisson computed from Fresnel’s integrals that the centre of the shadow of an opaque disc should be as bright as if the disc were absent. Arago tested this with a small metal disc and saw the spot. Fresnel won the prize in 1819. The spot is called the Poisson spot, the Arago spot, or the Fresnel spot. Joseph-Nicolas Delisle and Giacomo Filippo Maraldi had recorded it in the early eighteenth century without understanding it.

Calculation

A plane wave of wavelength \(\lambda\), \(k=2\pi/\lambda\), meets a screen in the plane \(z=0\); observe at distance \(z\) on the axis. In the Fresnel approximation, the field on axis due to an annular opening \(r_1<r<r_2\) is

\[ U=-\frac{ik}{z}\,e^{ikz}\int_{r_1}^{r_2}e^{ikr^2/2z}\,r\,dr=e^{ikz}\Big(e^{ikr_1^2/2z}-e^{ikr_2^2/2z}\Big), \]

normalized so that the unobstructed wave is \(e^{ikz}\).

Proposition

Behind an opaque disc of radius \(a\), in the Fresnel approximation, the on-axis intensity equals the unobstructed intensity \(|U_0|^2\), at every distance \(z\).

Derivation. By Babinet’s principle, the disc field is the free field minus the field through a circular hole of radius \(a\): \(U=e^{ikz}-e^{ikz}\big(1-e^{ika^2/2z}\big)=e^{ikz}e^{ika^2/2z}\). Its modulus is 1. Equivalently, take \(r_1=a,\ r_2\to\infty\) with a slowly decaying convergence factor. Off axis the field involves the Bessel function \(J_0(kr\rho/z)\), which the hero image computes. The spot has radius of order \(\lambda z/a\), so it is visible only for small, very round discs.

Exercise 9.1Use the annulus formula to find the on-axis intensity behind a circular hole of radius \(a\). For which \(z\) is it zero, and for which is it four times the incident intensity?

Show solution
With \(r_1=0,\ r_2=a\): \(|U|^2=|1-e^{i\pi N}|^2=4\sin^2(\pi N/2)\), where \(N=a^2/(\lambda z)\) is the Fresnel number. It is \(0\) when \(N\) is even and \(4\) when \(N\) is odd: an even or odd number of Fresnel zones is uncovered.

Exercise 9.2Arago’s disc had a diameter of about 2 mm. With \(\lambda=500\) nm and \(z=1\) m, estimate the radius of the spot, using the first zero of \(J_0\) at 2.405.

Show solution
Near the axis the intensity is proportional to \(J_0(ka\rho/z)^2\). The first zero is at \(\rho=\frac{2.405\,z}{ka}=\frac{2.405\times1}{(2\pi/5\times10^{-7})\times10^{-3}}\approx0.19\) mm. The spot is a fraction of a millimetre across, which is why the disc edge must be smooth to within a small fraction of a Fresnel zone.
Lesson 10

Probability and the law of large numbers

For the general reader. How many letters go astray each day, how many soldiers are killed by horse kicks each year, how many atoms decay each second? When an event is rare but has many chances to happen, its count follows the same law. Poisson derived it in a book about the reliability of jury verdicts, and he gave the phrase “law of large numbers” to probability.

Goals
  • Derive the Poisson distribution as a limit of binomials.
  • State Poisson’s law of large numbers for unequal probabilities.
  • Apply both to data and to juries.
Theorem (Poisson limit, 1837)

Let \(X_n\sim\mathrm{Bin}(n,p_n)\) with \(np_n\to\lambda>0\). Then for each \(k=0,1,2,\dots\)

\[ \Pr(X_n=k)\longrightarrow e^{-\lambda}\frac{\lambda^k}{k!}. \]

Proof. \(\binom nk p_n^k(1-p_n)^{n-k}=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{(np_n)^k}{k!}\cdot(1-p_n)^{n}(1-p_n)^{-k}\). The first factor tends to 1, the second to \(\lambda^k/k!\), the third to \(e^{-\lambda}\) because \(n\log(1-p_n)=-np_n+O(np_n^2)\to-\lambda\), and the last to 1. The distribution appeared in the Recherches sur la probabilité des jugements en matière criminelle et en matière civile (1837), with an earlier form in an 1830 paper on birth statistics. Abraham de Moivre had met the same expression in 1711 without singling it out. Its mean and variance both equal \(\lambda\).

Poisson’s law of large numbers

Let \(A\) occur in trial \(i\) with probability \(p_i\), independently, and let \(S_n\) be the number of occurrences in \(n\) trials. Then for every \(\varepsilon>0\)

\[ \Pr\left(\left|\frac{S_n}{n}-\frac{p_1+\cdots+p_n}{n}\right|>\varepsilon\right)\longrightarrow0. \]

Jacob Bernoulli had proved the case of equal \(p_i\) in 1713. Poisson’s argument was not rigorous by later standards. Pafnuty Chebyshev gave a short complete proof in 1846 using what is now Chebyshev’s inequality: \(\mathrm{Var}(S_n/n)=\frac1{n^2}\sum p_i(1-p_i)\le\frac1{4n}\).

Juries

Poisson modelled a juror as an independent judge who is right with probability \(p\), estimated \(p\) and the prior probability of guilt from French conviction statistics, and asked how the size of juries and the majority required affected wrongful convictions. The question was live, because French law on jury majorities changed during these years. Critics objected that the model treated moral judgements as games of chance.

Exercise 10.1Ladislaus Bortkiewicz (1898) recorded deaths by horse kick in Prussian cavalry corps: over 200 corps-years there were 109 years with 0 deaths, 65 with 1, 22 with 2, 3 with 3 and 1 with 4. Estimate \(\lambda\) and compare with the Poisson prediction.

Show solution
Total deaths \(=65+44+9+4=122\), so \(\lambda=122/200=0.61\). Expected counts \(200e^{-0.61}0.61^k/k!\): \(108.7,\ 66.3,\ 20.2,\ 4.1,\ 0.6\). The fit to \(109,65,22,3,1\) is very close.

Exercise 10.2Show that \(\mathbb E X=\mathrm{Var}X=\lambda\) for \(X\sim\mathrm{Poisson}(\lambda)\), and that the sum of independent Poisson variables with parameters \(\lambda,\mu\) is Poisson with parameter \(\lambda+\mu\).

Show solution
The generating function is \(\mathbb E s^X=\sum e^{-\lambda}\frac{(\lambda s)^k}{k!}=e^{\lambda(s-1)}\). Differentiating at \(s=1\): \(\mathbb EX=\lambda\), \(\mathbb E X(X-1)=\lambda^2\), so \(\mathrm{Var}X=\lambda^2+\lambda-\lambda^2=\lambda\). For the sum, the generating functions multiply: \(e^{\lambda(s-1)}e^{\mu(s-1)}=e^{(\lambda+\mu)(s-1)}\).

Exercise 10.3A jury of 12 convicts by a majority of at least 7. Each juror independently reaches the correct verdict with probability \(p\). Find the probability that the jury is correct for \(p=0.7\) and \(p=0.6\).

Show solution
\(\Pr=\sum_{k=7}^{12}\binom{12}{k}p^k(1-p)^{12-k}\). Numerically this is about \(0.882\) for \(p=0.7\) and \(0.665\) for \(p=0.6\). Accuracy depends strongly on the individual \(p\), which is why Poisson needed to estimate it from data.
Part III · Life and legacy
Lesson 11

The establishment of French science, 1816–1840

After Laplace, Poisson was the most powerful mathematician in France. He examined every candidate for the École Polytechnique, served as a councillor of the University from 1820, sat on Academy commissions, and in 1827 succeeded Laplace as geometer of the Bureau des Longitudes. He married Nancy de Bardi in 1817. He received the title of baron, which he never used.

He worked on one problem at a time and kept notes of future topics in a small portfolio, according to Guglielmo Libri. Arago reported that Poisson considered life good for only two things: doing mathematics and teaching it.

Rivalries

His relationship with Fourier was difficult. Poisson criticized the rigour of Fourier’s heat theory and then published his own treatments of heat, starting in 1815. Fourier wrote that Poisson “has too much talent to apply it to the work of others.” Poisson’s major treatise on heat appeared in 1835, after Fourier’s death. He also competed with Cauchy over elasticity and wave propagation, and with Claude-Louis Navier, whose equations of fluid motion Poisson derived independently in 1831.

The Galois report

In 1831 Poisson and Sylvestre François Lacroix reported to the Academy on the memoir in which Évariste Galois set out his theory of solvability of equations by radicals. They found the arguments too unclear to verify, but encouraged the author to develop and complete the work. Galois, embittered, died the next year. The memoir was published by Joseph Liouville in 1846, and the report is often cited as a famous misjudgement, though the referees’ complaint of obscurity was not unreasonable for what they had in hand.

Students

Generations of polytechniciens passed through his examinations and lectures. Among those he taught or supervised were Michel Chasles, Mikhail Ostrogradsky, Joseph Liouville and Sadi Carnot. Peter Gustav Lejeune Dirichlet, in Paris in the 1820s, knew him and his work.

DiscussionPoisson’s molecular theory of elasticity gave a wrong value of \(\nu\), and his objections to Fresnel and Galois look poor in hindsight. Yet his equations survive. What does his career suggest about the relationship between physical assumptions and mathematical method?

Lesson 12

Last years and legacy

In the 1830s Poisson set out to gather his work into a unified treatise on mathematical physics. The volumes that appeared were the Nouvelle théorie de l’action capillaire (1831), the enlarged second edition of the Traité de mécanique (1833), the Théorie mathématique de la chaleur (1835) and the Recherches sur la probabilité des jugements (1837). In 1837 he was made a peer of France as a representative of science. He had become a Fellow of the Royal Society in 1818 and received its Copley Medal in 1832.

He died at Sceaux, near Paris, on 25 April 1840, aged 58, and was buried at Père-Lachaise cemetery. His name is among the 72 inscribed on the Eiffel Tower.

Legacy

  • Mechanics: the Poisson bracket structures Hamiltonian mechanics, symplectic geometry and, through Dirac, canonical quantization. Poisson manifolds generalize it.
  • Potential theory: \(\nabla^2\phi=4\pi G\rho\) is Newtonian gravity in field form; numerical “Poisson solvers” are at the core of computational physics.
  • Analysis: the Poisson kernel and summation formula are central in harmonic analysis, number theory and signal processing.
  • Probability: the Poisson distribution and the Poisson process model rare events, arrivals and radioactive decay.
  • Physics: Poisson’s ratio in elasticity, the adiabatic law, and the spot that helped establish the wave theory of light.

Final projectChoose one object named after Poisson. Write a two-page account for a strong undergraduate that states it precisely, explains the problem Poisson was solving when he found it, and describes one modern use.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Siméon Denis Poisson, MacTutor History of Mathematics.
  2. Siméon Denis Poisson, Wikipedia; Siméon Denis Poisson, French Wikipedia.
  3. Siméon Denis Poisson, Mathematics Genealogy Project.
  4. M. Métivier, P. Costabel and P. Dugac (eds.), Siméon-Denis Poisson et la science de son temps, École Polytechnique, 1981.
  5. Y. Kosmann-Schwarzbach (ed.), Siméon-Denis Poisson: les mathématiques au service de la science, Éditions de l’École Polytechnique, 2013.

Mathematics and physics

  1. S. D. Poisson, Sur la variation des constantes arbitraires dans les questions de mécanique, J. École Polytechnique 8 (1809).
  2. S. D. Poisson, Remarques sur une équation qui se présente dans la théorie des attractions des sphéroïdes, Bull. Soc. Philomatique 3 (1813).
  3. S. D. Poisson, Traité de mécanique, 2nd ed., Bachelier, 1833.
  4. S. D. Poisson, Recherches sur la probabilité des jugements en matière criminelle et en matière civile, Bachelier, 1837.
  5. Arago spot, Wikipedia (history and references).
  6. I. Grattan-Guinness, Convolutions in French Mathematics, 1800–1840, Birkhäuser, 1990.
  7. S. M. Stigler, The History of Statistics, Harvard University Press, 1986, ch. 5.
  8. L. C. Evans, Partial Differential Equations, AMS, 2nd ed., 2010, ch. 2 (Poisson’s equation, Poisson kernel, wave equation).